Leetcode-95.Unique Binary Search Trees II(a representative recursion problem)

题目：

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,

Given n = 3, your program should return all 5 unique BST's shown below.

1         3     3      2      1
\       /     /      / \      \
3     2     1      1   3      2
/     /       \                 \
2     1         2                 3

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AC代码：
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
vector<TreeNode *>result;
if (n == 0)return result;
result = generateSubTree(1,n);
return result;
}

vector<TreeNode *> generateSubTree(int start,int end){//产生从start到end之间编号的子树
vector<TreeNode *>re;
if (end < start)return re;
if (end - start == 0){
TreeNode *temp = new TreeNode(end);
re.push_back(temp);
return re;
}
for (int root = start; root <= end;root++){//根节点编号
vector<TreeNode *>left = generateSubTree(start,root-1);//产生左子树
vector<TreeNode *>right = generateSubTree(root+1,end);//产生右子树
int leftnum = left.size();
int rightnum = right.size();
for (int i = 0; i < (leftnum==0?1:leftnum);i++){
for (int j = 0; j < (rightnum==0?1:rightnum);j++){
TreeNode *tem = new TreeNode(root);
tem->left = (leftnum==0?NULL:left[i]);
tem->right = (rightnum==0?NULL:right[j]);
re.push_back(tem);
}
}
}
return re;
}
};

分析：这道题是Leetcode-95.Unique Binary Search Trees的升级，基本思想是类似的，但是这里需要注意的是vector<TreeNode*>存放的是每棵树的根节点
的指针，容器的大小代表其存放的二叉搜索树的数量。在递归求解中，

1.将每个节点的指针依次存放到根节点的左右子树区域(left和right)。

2.通过数字root来划分左右字数的数字，并将它们形成的子二叉搜索树链接起来。。。